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Drop a perpendicular from $A$ to $BD$ that meets $BD$ at $E$ and join $CE$ . Since $\angle AED = 90^\circ$ and $\angle ADE = 60^\circ$ , $AED$ is a 30-60-90 triangle so $DE = \frac{1}{2}AD = \frac{1}{2}(2DC) = DC$ so $DCE$ is an isosceles triangle. We proceed to calculate various other angles and we use the above result in the third calculation.

$\angle DAE = 180^\circ - \angle ADE - \angle AED = 180^\circ - 60^\circ - 90^\circ = 30^\circ$ $\angle CDE = 180^\circ - \angle ADE = 180^\circ - 60^\circ = 120^\circ$ $\angle DCE = \angle DEC = \frac{1}{2}(180^\circ - \angle CDE) = \frac{1}{2}(180^\circ - 120^\circ) = 30^\circ$ $\angle ECB = \angle DCB - \angle DCE = 45^\circ - 30^\circ = 15^\circ$ $\angle EBC = 180^\circ - \angle DCB - \angle CDB = 180^\circ - 45^\circ - 120^\circ = 15^\circ$

Note from the above calculations that $\angle DAE = \angle DCE = 30^\circ$ and $\angle ECB = \angle EBC = 15^\circ$ from which we can deduce respectively that $EAC$ and $EBC$ are isosceles triangles giving us $EA = EC$ and $EC = EB$ respectively.

This gives us $EA = EB$ so $EAB$ must also be isosceles. Since $\angle AEB = 90^\circ$ , $AEB$ must be a 45-45-90 triangle and in conclusion we get $\angle CAB = \angle EAC + \angle EAB = 30^\circ + 45^\circ = 75^\circ$ which is our desired answer.

Given - ABC is a triangle in which angle ACB = 45 degrees . D is a point on AC such that AD = 2DC . Angle ADB = 60 degrees . To Find :-- Angle CAB. Construction : - Draw BE perpendicular to AC to meet AC at E .

In the figure , let DC = $x$ then AD = 2 $x$

Now, in right triangle BED , $\frac{BE}{DE}$ = tan 60
=> $\frac{BE}{DE}$ = $\frac{√(3)}{1}$ So, let BE = √(3) $y$ and DE = $y$

Now, in right triangle BEC , $\frac{BE}{CE}$ = tan 45 => $\frac{BE}{CE}$ = 1 .'. CE = BE = √(3) $y$
But CE = CD + DE = $x$ + $y$

.'. √(3) $y$ = $x$ + $y$ => $x$ = {√3 - 1} $y$

Now, AE = AD - DE = $2x$ - $y$ = 2{√3 - 1} $y$ - $y$ = 2√3 $y$ - 2 $y$ - $y$ = {2√3 - 3} $y$

Consider right triangle ABE , tan CAB = $\frac{BE}{AE}$ => tan ∠CAB = $\frac{√3y}{(2√3 - 3)y}$ => tan ∠CAB = 3.732........... => tan ∠CAB = tan 75
=> ∠CAB= 75 ---------------------------------------------------------an easy solution by Vishwash Kumar

From the law of sine, $\frac {a} {\sin120^{\circ}}=\frac {x} {\sin15^{\circ}}$ $\Rightarrow a=\frac{\sin120^{\circ}}{\sin15^{\circ}}x=\frac{\sin60^{\circ}}{\sin15^{\circ}}x=\frac{\frac{\sqrt3}{2}}{\frac{1}{\sqrt2(1+\sqrt3)}} x=\frac{\sqrt6(1+\sqrt3)}{2}x$ $\color{teal} {\left[\text {Used }\sin15^{\circ} =\sin(45^{\circ}-30^{\circ}) \text{ to get, } \sin15^{\circ}=\frac{1}{\sqrt2(1+\sqrt3)}\right] }$

From the law of cosine, $c=\sqrt{a^2+(3x)^2-2\times a\times 3x\times \cos 45^{\circ}}=\sqrt{6x^2}=\sqrt6x$

Using the law of sine again we get, $\frac {c} {\sin 45^{\circ}}=\frac {a} {\sin \theta}$ $\Rightarrow \sin \theta =\frac{a} {c} \sin 45^{\circ}=\frac{\frac{\sqrt6(1+\sqrt3)}{2}x} {\sqrt6x}\times \frac{1}{\sqrt2}=\frac{1+\sqrt3}{2\sqrt2}$ $\Rightarrow \theta =\sin^{-1}\left(\frac{1+\sqrt3}{2\sqrt2}\right) =75^{\circ}\, \text{or,} \, 105^{\circ}\, \color{teal} {[\text{As, } 0^{\circ}<\theta<180^{\circ}]}$ Now, $\because AC>AB$ $\Rightarrow \angle B>\angle C=45^{\circ}$ $\Rightarrow \angle B+\angle C>90^{\circ}$ $\Rightarrow \angle A<(180^{\circ}-90^{\circ})$ $\Rightarrow \theta<90^{\circ}$

Thus, only possible value of, $\theta =\color{#0C6AC7} {\boxed {75^{\circ}}}$

Solution 1: Let $O$ be the circumcenter of triangle $CBD$ . Since $\angle BDC$ is obtuse, $O$ and $D$ lie on the opposite sides of $BC$ . Then $\angle BOD = 2 \angle BCD = 90^\circ, \angle DOC = 2 \angle CBD = 30^\circ$ and thus, $\angle BOC = \angle BOD + \angle BOC = 120^\circ$ .

In isosceles triangle $OBC$ , we have $\angle OCB=\angle OBC = \frac{180^\circ - \angle BOC}{2} = 30^\circ.$ In isosceles triangle $ODC$ , we have $\angle ODC = \angle OCD = \angle DCB + \angle BCO = 45^\circ + 30^\circ = 75^\circ$ and it follows that $\angle DOC = 30^\circ$ .

Let $E$ be the intersection point between lines $BC$ and $DO$ . Since $\angle BCO = \angle ECO= 30^\circ = \angle DOC = \angle EOC$ , therefore, $EOC$ is an isosceles triangle, so $EO=CE$ . We have that, $EOB$ is a right triangle at $O$ , thus, $EB = \frac {EO}{\sin 30^\circ} = 2 EC$ . From the ratio of lengths, $CD:DA = 1:2 = CE:EB$ , thus, $DE \parallel AB$ . Hence, $\angle ABD = \angle BDO = \angle DBO$ $=\frac{180^\circ - \angle DOB}{2} = \frac{180^\circ - 90^\circ}{2} = 45^\circ$ . Therefore, $\angle CAB = 180^\circ - \angle ADB - \angle ABD = 180^\circ - 60^\circ -45^\circ = 75^\circ$ .

Solution 2: Let $\angle CAB = \alpha$ . Since $\angle ADB = 60^\circ$ , $0 < \alpha < 120^\circ$ . Using the sine rule on triangle $DBC$ , we get $\frac {DB}{\sin 45^\circ} = \frac {DC} {\sin 15^\circ}$ . Using the sine rule on triangle $DBA$ , we get $\frac {DB}{\sin \alpha } = \frac {DA} {\sin (120^\circ - \alpha) }$ . This gives $\frac {\sin (120^\circ - \alpha)} {\sin \alpha} = \frac {DA}{DB} = \frac {2DC}{DB} = \frac { 2 \sin 15^\circ} {\sin45^\circ}$ .

Expanding out the LHS of the above equation, we get $\frac {\sin (120^\circ - \alpha)} {\sin \alpha } = \frac { \sin 120^\circ \cos \alpha - \cos 120^\circ \sin \alpha } {\sin \alpha} = \sin 120^\circ \cot \alpha - \cos 120^\circ$ . Since $\cot \alpha$ is strictly decreasing in the interval $0 < \alpha < 120^\circ$ , the LHS will be strictly decreasing in the same interval. We see that the LHS decrease from $\infty$ (when $\alpha = 0^\circ$ to $0$ (when $\alpha = 120^\circ$ ), hence there exists a unique solution to $\frac {\sin (120^\circ - \alpha)} {\sin \alpha} = \frac { 2 \sin 15^\circ} {\sin45^\circ}$ .

We will now show that $\alpha = 75^\circ$ satisfies the equation. We have $2 \sin 75^\circ \sin 15^\circ = \cos (75^\circ + 15^\circ) - \cos(75^\circ - 15^\circ) = \frac {1}{2} = \sin^2 45^\circ$ . Therefore, $\frac{2\sin 15^\circ}{\sin 45^\circ} = \frac{\sin 45^\circ}{\sin 75^\circ}$ $= \frac { \sin (120^\circ -\alpha) } {\sin \alpha}$ . Thus, the answer is 75.